Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9245 Accepted Submission(s): 2327 

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

 

Output

For each test case, you should output the rightmost digit of N^N.

 

Sample Input

234

 

Sample Output

76
Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
#include<stdio.h>main(){    int n,cas,m;    scanf("%d",&cas);    while(cas--)    {        scanf("%d",&n);        m=n%10;        if(m==0||m==1||m==5||m==6)            printf("%d\n",m);        if(m==2)        {            if (n%4==1) printf("2\n");            if (n%4==2) printf("4\n");            if (n%4==3) printf("8\n");            if (n%4==0) printf("6\n");        }        if(m==3)        {            if (n%4==1) printf("3\n");            if (n%4==2) printf("9\n");            if (n%4==3) printf("7\n");            if (n%4==0) printf("1\n");        }        if(m==7)        {            if (n%4==1) printf("7\n");            if (n%4==2) printf("9\n");            if (n%4==3) printf("3\n");            if (n%4==0) printf("1\n");        }        if(m==8)        {            if (n%4==1) printf("8\n");            if (n%4==2) printf("4\n");            if (n%4==3) printf("2\n");            if (n%4==0) printf("6\n");        }        if (m==4)        {            if (n%2==1) printf("4\n");            if (n%2==0) printf("6\n");        }        if (m==9)        {            if (n%2==1) printf("9\n");            if (n%2==0) printf("1\n");        }    }}